A bullet fired at an angle of ` 30^(@)` with the horizontal hits the ground ` 3 sqrt 3 km ` away. Can wr hit a target at a distance of `6 sqrt 2km ` by adjustion its angle of projection?

A bullet fired at an angle of ` 30^(@)` with the horizontal hits the g

1.	A bullet fired at an angle of ` 30^(@)` with the horizontal hits the ground ` 3 sqrt 3 km ` away. Can wr hit a target at a distance of `6 sqrt 2km ` by adjustion its angle of projection?
Answer» Here, `theta =30^(@)`, `R=3 sqrt 3 km =300 sqrt3 m` As, `R=(u^(2) sin 2 theta)/g` :. 3000 sqrt 3 =(u^(2) sin 2 xx 30^(@))/g` or `u^(2)/g =(3-00 sqrt 3)/(sin 50^(@)) =(3000sqrt 3)/(sqer3//2)` `=6000 m =6 km ` The maximum horizontal rage `=u^(2)/g =6 km`, which is less than `6 sqrt2 km`. Thus the bullet will never hit the target at a distance `6 sqrt 2 km ` by adjusting anlge of projection.

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A bullet fired at an angle of ` 30^(@)` with the horizontal hits the ground ` 3 sqrt 3 km ` away. Can wr hit a target at a distance of `6 sqrt 2km ` by adjustion its angle of projection?

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